Solve the following system of equations, algebraically, by substitution.
y + 2x = 3 x = 3y + 5
y + 2x = 3
x = 3y + 5 y + 2x = 3 {first equation} y + 2(3y + 5) = 3 {substituted 3y + 5, in for x, into first equation} y + 6y + 10 = 3 {used distributive property} 7y + 10 = 3 {combined like terms} 7y = 7 {subtracted 10 from each side} y = 1 {divided each side by 7} x = 3y + 5 {second equation} x = 3(1) + 5 {substituted 1, in for x, into second equation} x = 3 + 5 {multiplied} x = 2 {added} (2,1) is the solution to the system Using the graphing calculator, you can see (2,1) is the point of intersection of the two lines, as shown below: Comments are closed.

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